Is there a way to visualize how taking the logarithm of a function doesn't change the locations of the max and min?

Question

I was attempting to take the second derivative of the binomial distribution with respect to p to show that p = j/n is a maximum (where the binomial coefficients are $\binom{n}{j}$). From this answer it seems like the easiest way is to take the logarithm of the binomial distribution equation first and then taking the second derivative of that equation (and prove that it is less than zero at p = j/n). My knowledge of logarithms is weak and I had trouble visualising how taking the logarithm of a function would not change the location of the maximum and minimum (and also the concavity?).

Answer 1

This is nothing special to do with the logarithm function. Given any strictly increasing function, the locations of maximums and minimums are going to be preserved.

To see this, let $f : \mathbb{R} \rightarrow \mathbb{R}$ be any function you like, and let $g : \mathbb{R} \rightarrow \mathbb{R}$ be any strictly increasing function. Then (essentially by definition) we have that $f(x) < f(y)$ if and only if $g(f(x)) < g(f(y))$.

(By the same logic, strictly decreasing functions will always switch the order of maximums and minimums around).

Answer 2

This holds not just for logarithms, but any monotonically increasing function. Let's say $f(x)$ is your function with maxima $x_0$. Then -

$$f(x_0)>f(x_0+\epsilon)$$

and

$$f(x_0)>f(x_0-\epsilon)$$

Now, take any monotonically increasing function $g(x)$. This means that if $x_1>x_2$ then $g(x_1)>g(x_2)$.

Apply this to the two equations above and you see how a maxima remains a maxima.